Question: $3x^2-x^2y+y^4=1$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2xy-6x}{4y^3+x^2}$ (Choice B) B $\dfrac{xy}{3x+2y^3}$ (Choice C) C $\dfrac{2xy-6x}{4y^3-x^2}$ (Choice D) D $\dfrac{6x+4y^3}{2xy}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $3x^2-x^2y+y^4=1$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 3x^2-x^2y+y^4&=1 \\\\ \dfrac{d}{dx}(3x^2-x^2y+y^4)&=\dfrac{d}{dx}(1) \\\\ \dfrac{d}{dx}(3x^2)-\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(y^4)&=0 \\\\ 6x-\Bigl(2x\cdot y+x^2\cdot\dfrac{dy}{dx}\Bigr)+4y^3\cdot\dfrac{dy}{dx}&=0 \\\\ 6x-2xy-x^2\cdot\dfrac{dy}{dx}+4y^3\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 6x-2xy-x^2\cdot\dfrac{dy}{dx}+4y^3\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(4y^3-x^2)&=2xy-6x \\\\ \dfrac{dy}{dx}&=\dfrac{2xy-6x}{4y^3-x^2} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{2xy-6x}{4y^3-x^2}$.